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Determination of singular time-dependent coefficients
for wave equations from full and partial data
Guanghui Hu, Yavar Kian
To cite this version:
Guanghui Hu, Yavar Kian. Determination of singular time-dependent coefficients for wave equations
from full and partial data. Inverse Problems and Imaging , AIMS American Institute of Mathematical
Sciences, 2018, 12 (3), pp.745-772. �10.3934/ipi.2018032�. �hal-01544811�
DETERMINATION OF SINGULAR TIME-DEPENDENT COEFFICIENTS FOR WAVE EQUATIONS FROM FULL AND PARTIAL DATA
GUANGHUI HU AND YAVAR KIAN
Abstract. We study the problem of determining uniquely a time-dependent singular potential q, appearing in the wave equation ∂2
tu − ∆xu + q(t, x)u = 0 in Q = (0, T ) × Ω with T > 0 and Ω a C2bounded domain of
Rn, n> 2. We start by considering the unique determination of some singular time-dependent coefficients from observations on ∂Q. Then, by weakening the singularities of the set of admissible coefficients, we manage to reduce the set of data that still guaranties unique recovery of such a coefficient. To our best knowledge, this paper is the first claiming unique determination of unbounded time-dependent coefficients, which is motivated by the problem of determining general nonlinear terms appearing in nonlinear wave equations.
Keywords: Inverse problems, wave equation, time dependent coefficient, singular coefficients, Carleman estimate.
Mathematics subject classification 2010 : 35R30, 35L05.
1. Introduction 1.1. Statement of the problem. Let Ω be a C2
bounded domain of Rn, n
> 2, and fix Σ = (0, T ) × ∂Ω, Q = (0, T ) × Ω with 0 < T < ∞. We consider the wave equation
∂t2u − ∆xu + q(t, x)u = 0, (t, x) ∈ Q, (1.1)
where the potential q is assumed to be an unbounded real valued coefficient. In this paper we seek unique determination of q from observations of solutions of (1.1) on ∂Q.
1.2. Obstruction to uniqueness and set of full data for our problem. Let ν be the outward unit normal vector to ∂Ω, ∂ν = ν · ∇xthe normal derivative and from now on let be the differential operators
:= ∂2
t− ∆x. It has been proved by [40], that, for T > Diam(Ω), the data
Aq= {(u|Σ, ∂νu|Σ) : u ∈ L2(Q), u + qu = 0, u|t=0= ∂tu|t=0= 0} (1.2)
determines uniquely a time-independent potential q. On the other hand, due to domain of dependence arguments, there is no hope to recover even smooth time-dependent coefficients restricted to the set
D = {(t, x) ∈ Q : t ∈ (0, Diam(Ω)/2) ∪ (T − Diam(Ω)/2, T ), dist(x, ∂Ω) > min(t, T − t)}
from the data Aq (see [32, Subsection 1.1]). Therefore, even when T is large, for the global recovery of general
time-dependent coefficients the information on the bottom t = 0 and the top t = T of Q are unavoidable. Thus, for our problem the extra information on {t = 0} and {t = T }, of solutions u of (1.1), can not be completely removed. In this context, we introduce the set of data
Cq = {(u|Σ, ut=0, ∂tu|t=0, ∂νu|Σ, u|t=T, ∂tu|t=T) : u ∈ L2(Q), u + qu = 0}.
and we recall that [25] proved that, for q ∈ L∞(Q), the data Cq determines uniquely q. From now on we
will refer to Cq as the set of full data for our problem and we mention that [31, 32, 33] proved recovery of
bounded time-dependent coefficients q from partial data corresponding to partial knowledge of the set Cq.
The goal of the present paper is to prove recovery of singular time-dependent coefficients q from full and partial data.
1.3. Physical and mathematical motivations. Physically speaking, our inverse problem consists of de-termining unstable properties such as some rough time evolving density of an inhomogeneous medium from disturbances generated on the boundary and at initial time, and measurements of the response. The goal is to determine the function q which describes the property of the medium. Moreover, singular time-dependent coefficients can be associated to some unstable time-evolving phenomenon that can not be modeled by bounded time-dependent coefficients or time independent coefficients.
Let us also observe that, according to [11, 27], for parabolic equations the recovery of nonlinear terms, appearing in some suitable nonlinear equations, can be reduced to the determination of time-dependent coefficients. In this context, the information that allows to recover the nonlinear term is transferred, throw a linearization process, to a time-dependent coefficient depending explicitly on some solutions of the nonlinear problem. In contrast to parabolic equations, due to the weak regularity of solutions, it is not clear that this process allows to transfer the recovery of nonlinear terms, appearing in a nonlinear wave equation, to a bounded time-dependent coefficient. Thus, in order to expect an application of the strategy set by [11, 27] to the recovery of nonlinear terms for nonlinear wave equations, it seems important to consider recovery of singular time-dependent coefficients.
1.4. known results. The problem of determining coefficients appearing in hyperbolic equations has at-tracted many attention over the last decades. This problem has been stated in terms of recovery of a time-independent potential q from the set Aq. For instance, [40] proved that Aq determines uniquely a
time-independent potential q, while [16] proved that partial boundary observations are sufficient for this problem. We recall also that [4, 5, 30, 44] studied the stability issue for this problem.
Several authors considered also the problem of determining time-dependent coefficients appearing in wave equations. In [43], the authors shown that the knowledge of scattering data determines uniquely a smooth time-dependent potential. In [41], the authors studied the recovery of a time-dependent potential q from data on the boundary ∂Ω for all time given by (u|R×∂Ω, ∂νu|R×∂Ω) of forward solutions of (1.1) on the
infinite time-space cylindrical domain Rt× Ω instead of Q. As for [39], the authors considered this problem
at finite time on Q and they proved the recovery of q restricted to some strict subset of Q from Aq. Isakov
established in [25, Theorem 4.2] unique global determination of general time-dependent potentials on the whole domain Q from the important set of full data Cq. By applying a result of unique continuation for wave
equation, which is valid only for coefficients analytic with respect to the time variable (see for instance the counterexample of [1]), [17] proved unique recovery of time-dependent coefficients from partial knowledge of the data Aq. In [42], the author extended the result of [41]. Moreover, [46] established the stable recovery
of X-ray transforms of time-dependent potentials and [2, 6] proved log-type stability in the determination of time-dependent coefficients with data similar to [25] and [39]. In [31, 32, 33], the author proved uniqueness and stability in the recovery of several time-dependent coefficients from partial knowledge of the full set of data Cq. It seems that the results of [31, 32, 33] are stated with the weakest conditions so far that allows to
recover general bounded time-dependent coefficients. More recently, [34] proved unique determination of such coefficients on Riemannian manifolds. We mention also the work of [45] who determined some information about time-dependent coefficients from the Dirichlet-to-Neumann map on a cylinder-like Lorentzian manifold related to the wave equation. We refer to the work [10, 12, 20, 21, 35] for determination of time-dependent coefficients for fractional diffusion, parabolic and Schrödinger equations have been considered.
In all the above mentioned results, the authors considered time-dependent coefficients that are at least bounded. There have been several works dealing with recovery of non-smooth coefficients appearing in elliptic equations such as [9, 15, 19, 23]. Nevertheless, to our best knowledge, except the present paper, there is no work in the mathematical literature dealing with the recovery of singular time-dependent coefficients q even from the important set of full data Cq.
1.5. Main results. The main purpose of this paper is to prove the unique global determination of time-dependent and unbounded coefficient q from partial knowledge of the observation of solutions on ∂Q = ({0} × Ω) ∪ Σ ∪ ({T } × Ω). More precisely, we would like to prove unique recovery of unbounded coefficient
q ∈ Lp1(0, T ; Lp2(Ω)), p
1 > 2, p2 > n, from partial knowledge of the full set of data Cq. We start by
considering the recovery of some general unbounded coefficient q from restriction of Cq on the bottom t = 0
and top t = T of the time-space cylindrical domain Q. More precisely, for q ∈ Lp1(0, T ; Lp2(Ω)), p
1 > 2,
p2> n, we consider the recovery of q from the set of data
Cq(0) = {(u|Σ, ∂tu|t=0, ∂νu, u|t=T, ∂tu|t=T) : u ∈ K(Q), u + qu = 0, u|t=0= 0},
or the set of data
Cq(T ) = {(u|Σ, ut=0, ∂tu|t=0, ∂νu, u|t=T) : u ∈ K(Q), u + qu = 0},
where K(Q) = C([0, T ]; H1(Ω)) ∩ C1([0, T ]; L2(Ω)). In addition, assuming that T > Diam(Ω), we prove the recovery of q from the set of data
Cq(0, T ) = {(u|Σ, ∂tu|t=0, ∂νu, u|t=T) : u ∈ K(Q), u + qu = 0, u|t=0= 0}.
Our first main result can be stated as follows
Theorem 1.1. Let p1 ∈ (2, +∞), p2 ∈ (n, +∞) and let q1, q2 ∈ Lp1(0, T ; Lp2(Ω)). Then, either of the
following conditions:
Cq1(0) = Cq2(0), (1.3)
Cq1(T ) = Cq2(T ), (1.4)
implies that q1= q2. Moreover, assuming that T > Diam(Ω), the condition
Cq1(0, T ) = Cq2(0, T ) (1.5)
implies that q1= q2.
We consider also the recovery of a time-dependent and unbounded coefficient q from restriction of the data Cq on the lateral boundary Σ. Namely, for all ω ∈ Sn−1 = {x ∈ Rn : |x| = 1} we introduce the
ω-shadowed and ω-illuminated faces
∂Ω+,ω= {x ∈ ∂Ω : ν(x) · ω > 0}, ∂Ω−,ω = {x ∈ ∂Ω : ν(x) · ω 6 0}
of ∂Ω. Here, for all k ∈ N∗, · denotes the scalar product in Rk given by
x · y = x1y1+ . . . + xkyk, x = (x1, . . . , xk) ∈ Rk, y = (y1, . . . , yk) ∈ Rk.
We define also the parts of the lateral boundary Σ taking the form Σ±,ω = (0, T ) × ∂Ω±,ω. We fix ω0∈ Sn−1
and we consider V = (0, T ) × V0 with V0 a closed neighborhood of ∂Ω−,ω0 in ∂Ω. Then, we study the
recovery of q ∈ Lp(Q), p > n + 1, from the data
Cq(T, V ) = {(u|Σ, u|t=0, ∂tu|t=0, ∂νu|V, u|t=T) : u ∈ H1(Q), u + qu = 0}
and the determination of a time-dependent coefficient q ∈ L∞(0, T ; Lp(Ω)), p > n, from the data
Cq(0, T, V ) = {(u|Σ, ∂tu|t=0, ∂νu|V, u|t=T) : u ∈ L2(0, T ; H1(Ω)), u + qu = 0, u|t=0= 0}.
We refer to Section 2 for the definition of this set. Our main result can be stated as follows. Theorem 1.2. Let p ∈ (n + 1, +∞) and let q1, q2∈ Lp(Q). Then, the condition
Cq1(T, V ) = Cq2(T, V ) (1.6)
implies that q1= q2.
Theorem 1.3. Let p ∈ (n, +∞) and let q1, q2∈ L∞(0, T ; Lp(Ω)). Then, the condition
Cq1(0, T, V ) = Cq2(0, T, V ) (1.7)
To our best knowledge the result of Theorem 1.1, 1.2 and 1.3 are the first results claiming unique determination of unbounded time-dependent coefficients for the wave equation. In Theorem 1.1, we prove recovery of coefficients q, that can admit some singularities, by making restriction on the set of full data Cq on the bottom t = 0 and the top t = T of Q. While, in Theorem 1.2 and 1.3, we consider less singular
time-dependent coefficients, in order to restrict the data on the lateral boundary Σ = (0, T ) × ∂Ω.
We mention also that the uniqueness result of Thorem 1.3 is stated with data close to the one considered by [31, 32], where determination of bounded time-dependent potentials is proved with conditions that seems to be one of the weakest so far. More precisely, the only difference between [31, 32] and Theorem 1.3 comes from the restriction on the Dirichlet boundary condition ([31, 32] consider Dirichlet boundary condition supported on a neighborhood of the ω0-shadowed face, while in Theorem 1.3 we do not restrict the support
of the Dirichlet boundary).
In the present paper we consider two different approaches which depend mainly on the restriction that we make on the set of full data Cq. For Theorem 1.1, we use geometric optics solutions corresponding to
oscillating solutions of the form u(t, x) =
N
X
j=1
aj(t, x)eiλψj(t,x)+ Rλ(t, x), (t, x) ∈ Q, (1.8)
with λ > 1 a large parameter, Rλ a remainder term that admits a decay with respect to the parameter
λ and ψj, j = 1, .., N , real valued. For N = 1, these solutions correspond to a classical tool for proving
determination of time independent or time-dependent coefficients (e. g. [2, 3, 4, 6, 39, 41, 40]). In a similar way to [34], we consider in Theorem 1.1 solutions of the form (1.8) with N = 2 in order to be able to restrict the data at t = 0 and t = T while avoiding a "reflection". It seems that in the approach set so far for the construction of solutions of the form (1.8), the decay of the remainder term Rλ relies in an important
way to the fact that the coefficient q is bounded (or time independent). In this paper, we prove how this construction can be extended to unbounded time-dependent coefficients.
The approach used for Theorem 1.1 allows in a quite straightforward way to restrict the data on the bottom t = 0 and on the top t = T of Q. Nevertheless, it is not clear how one can extend this approach to restriction on the lateral boundary Σ without requiring additional smoothness or geometrical assumptions. For this reason, in order to consider restriction on Σ, we use a different approach where the oscillating solutions (1.8) are replaced by exponentially growing and exponentially decaying solutions of the form
u(t, x) = e±λ(t+x·ω)(a(t, x) + wλ(t, x)), (t, x) ∈ Q, (1.9)
where ω ∈ Sn−1 and wλ admits a decay with respect to the parameter λ. The idea of this approach, which
is inspired by [5, 31, 32, 33] (see also [8, 29] for elliptic equations), consists of combining results of density of products of solutions with Carleman estimates with linear weight in order to be able to restrict at the same time the data on the bottom t = 0, on the top t = T and on the lateral boundary Σ of Q. For the construction of these solutions, we use Carleman estimates in negative order Sobolev space. To our best knowledge this is the first extension of this approach to singular time-dependent coefficients.
1.6. Outline. This paper is organized as follows. In Section 2, we start with some preliminary results and we define the set of data Cq(0), Cq(T ), Cq(0, T ), Cq(T, V ) and Cq(0, T, V ). In Section 3, we prove Theorem
1.1 by mean of geometric optics solutions of the form (1.8). Then, Section 4 and Section 5 are respectively devoted to the proof of Theorem 1.2 and Theorem 1.3.
2. Preliminary results
In the present section we define the set of data Cq(T, V ), Cq(0, T, V ) and we recall some properties of
the solutions of (1.1) for any q ∈ Lp1(Q), with p
1 > n + 1, or, for q ∈ L∞(0, T ; Lp2(Ω)), with p2> n. For
this purpose, in a similar way to [32], we will introduce some preliminary tools. We define the space H(Q) = {u ∈ H1(Q) : u = (∂t2− ∆x)u ∈ L2(Q)},
H,∗(Q) = {u ∈ L2(0, T ; H1(Ω)) : u = (∂t2− ∆x)u ∈ L2(Q)},
with the norm
kuk2H (Q)= kuk 2 H1(Q)+ (∂t2− ∆x)u 2 L2(Q), kuk2H ,∗(Q)= kuk 2 L2(0,T ;H1(Ω))+ (∂t2− ∆x)u 2 L2(Q).
We consider also the space
S = {u ∈ H1(Q) : (∂t2− ∆x)u = 0} (resp S∗= {u ∈ L2(0, T ; H1(Ω)) : (∂t2− ∆x)u = 0})
and topologize it as a closed subset of H1(Q) (resp L2(0, T ; H1(Ω))). In view of [32, Proposition 4], the
maps
τ0w = (w|Σ, w|t=0, ∂tw|t=0), τ1w = (∂νw|Σ, w|t=T, ∂tw|t=T), w ∈ C∞(Q),
can be extended continuously to τ0: H,∗(Q) → H−3(0, T ; H−
1
2(∂Ω)) × H−2(Ω) × H−4(Ω), τ1: H,∗(Q) →
H−3(0, T ; H−32(∂Ω)) × H−2(Ω) × H−4(Ω). Here for all w ∈ C∞(Q) we set
τ0w = (τ0,1w, τ0,2w, τ0,3w), τ1w = (τ1,1w, τ1,2w, τ1,3w),
where
τ0,1w = w|Σ, τ0,2w = w|t=0, τ0,3w = ∂tw|t=0, τ1,1w = ∂νw|Σ, τ1,2w = w|t=T, τ1,3w = ∂tw|t=T.
Therefore, we can introduce
H = {τ0u : u ∈ H(Q)} ⊂ H−3(0, T ; H− 1 2(∂Ω)) × H−2(Ω) × H−4(Ω), H∗= {(τ0,1u, τ0,3u) : u ∈ H,∗(Q), τ0,2u = 0} ⊂ H−3(0, T ; H− 1 2(∂Ω)) × H−4(Ω).
By repeating the arguments used in [32, Proposition 1], one can check that the restriction of τ0 to S (resp
S∗) is one to one and onto. Thus, we can use (τ0|S)−1(resp (τ0|S∗)
−1) to define the norm of H (resp H ∗) by k(f, v0, v1)kH= (τ0|S)−1(f, v0, v1) H1(Q), (f, v0, v1) ∈ H, (resp k(f, v1)kH∗= (τ0|S∗) −1(f, 0, v 1) L2(0,T ;H1(Ω)), (f, v1) ∈ H∗).
Let us consider the initial boundary value problem (IBVP in short) ∂t2v − ∆xv + qv = F (t, x), (t, x) ∈ Q, v(0, x) = v0(x), ∂tv(0, x) = v1(x), x ∈ Ω v(t, x) = 0, (t, x) ∈ Σ. (2.1)
We have the following well-posedness result for this IBVP when q is unbounded.
Proposition 2.1. Let p1∈ (1, +∞) and p2∈ (n, +∞). For q ∈ Lp1(0, T ; Lp2(Ω)), v0∈ H01(Ω), v1∈ L2(Ω)
and F ∈ L2(Q), problem (2.1) admits a unique solution v ∈ C([0, T ]; H01(Ω)) ∩ C1([0, T ]; L2(Ω)) satisfying
kvkC([0,T ];H1
0(Ω))+ kvkC1([0,T ];L2(Ω))6 C(kv0kH1(Ω)+ kv1kL2(Ω)+ kF kL2(Q)), (2.2)
with C depending only on p1, p2, n, T , Ω and any M > kqkLp1(0,T ;Lp2(Ω)).
Proof. According to the second part of the proof of [37, Theorem 8.1, Chapter 3], [37, Remark 8.2, Chapter 3] and [37, Theorem 8.3, Chapter 3], the proof of this proposition will be completed if we show that for any v ∈ W2,∞(0, T ; H1
0(Ω)) solving (2.1) the a priori estimate (2.2) holds true. Without lost of generality we
assume that v is real valued. From now on we consider this estimate. We define the enery E(t) at time t ∈ [0, T ] by
E(t) := Z
Ω
|∂tv(t, x)|2+ |∇xv(t, x)|2 dx.
Multiplying (2.1) by ∂tv and integrating by parts we get
E(t) − E(0) = −2 Z t 0 Z Ω q(s, x)v(s, x)∂tv(s, x)dxds + 2 Z t 0 Z Ω F (s, x)∂tv(s, x)dxds. (2.3)
On the other hand, we have Z t 0 Z Ω q(s, x)v(s, x)∂tv(s, x)dxds 6 Z t 0 kqv(s, ·)kL2(Ω)k∂tv(s, ·)kL2(Ω)ds. (2.4)
Applying the Sobolev embedding theorem and the Hölder inequality, for all s ∈ (0, T ) we get kqv(s, ·)kL2(Ω)6 kq(s, ·)kLp2(Ω)kv(s, ·)k L 2p2 p2−2(Ω) 6 C kq(s, ·)kLp2(Ω)kv(s, ·)kHp2n(Ω) 6 C kq(s, ·)kLp2(Ω)kv(s, ·)kH1(Ω)
with C depending only on Ω. Then, the Poincarré inequality implies
kqv(s, ·)kL2(Ω)6 C kq(s, ·)kLp2(Ω)k∇xv(s, ·)kL2(Ω)6 C kq(s, ·)kLp2(Ω)E(s)
1 2,
where C depends only on Ω. Thus, from (2.4), we get Z t 0 Z Ω q(s, x)v(s, x)∂tv(s, x)dxds 6 C Z t 0 kq(s, ·)kLp2(Ω)E(s)ds 6 kqkLp1(0,T ;Lp2(Ω)) Z t 0 E(s)p1−1p1 ds p1−1p1 . (2.5) In the same way, an application of the Hölder inequality yields
Z t 0 Z Ω F (s, x)∂tv(s, x)dxds 6 T 1 2p1 kF k L2(Q) Z t 0 E(s)p1−1p1 ds p1−12p1 6 Tp11 kF k2 L2(Q)+ Z t 0 E(s)p1−1p1 p1−1p1 . Combining this estimate with (2.3)-(2.5), we deduce that
E(t) 6 E(0) + C kF k2L2(Q)+ C Z t 0 E(s)p1−1p1 p1−1p1 , where C depends only on T , Ω and any M > kqkLp1(0,T ;Lp2(Ω)). By taking the power
p1
p1−1 on both side of
this inequality, we get E(t)p1−1p1 6 C kv0kH1(Ω)+ kv1kL2(Ω)+ kF kL2(Q) p1−12p1 + C Z t 0 E(s)p1−1p1 ds. Then, the Gronwall inequality implies
E(t)p1−1p1 6 C kv0kH1(Ω)+ kv1kL2(Ω)+ kF kL2(Q) p1−12p1 eCt 6 Ckv0kH1(Ω)+ kv1kL2(Ω)+ kF kL2(Q) p1−12p1 eCT. From this last estimate one can easily deduce (2.2).
Let us introduce the IBVP
∂2 tu − ∆xu + q(t, x)u = 0, in Q, u(0, ·) = v0, ∂tu(0, ·) = v1, in Ω, u = g, on Σ. (2.6)
We are now in position to state existence and uniqueness of solutions of this IBVP for (g, v0, v1) ∈ H and
Proposition 2.2. Let (g, v0, v1) ∈ H, q ∈ Lp(Q), p > n + 1. Then, the IBVP (2.6) admits a unique weak
solution u ∈ H1(Q) satisfying
kukH1(Q)6 C k(g, v0, v1)kH (2.7)
and the boundary operator Bq : (g, v0, v1) 7→ (τ1,1u|V, τ1,2u) is a bounded operator from H to
H−3(0, T ; H−32(V0)) × H−2(Ω).
Proof. We split u into two terms u = v + τ0−1(g, v0, v1) where v solves
∂t2v − ∆xv + qv = qτ0−1(g, v0, v1), (t, x) ∈ Q, v|t=0= ∂tv|t=0 = 0, v|Σ = 0. (2.8)
Since q ∈ Lp(Q) and τ0−1(g, v0, v1) ∈ H1(Q), by the Sobolev embedding theorem we have qτ0−1(g, v0, v1) ∈
L2(Q). Thus, according to Proposition 2.1 one can check that the IBVP (2.8) has a unique solution v ∈ C1([0, T ]; L2(Ω)) ∩ C([0, T ]; H1 0(Ω)) satisfying kvkC1([0,T ];L2(Ω))+ kvkC([0,T ];H1 0(Ω))6 C qτ0−1(g, v0, v1) L2(Q) 6 C kqkLp(Q) τ0−1(g, v0, v1) H1(Q). (2.9)
Thus, u = v+τ0−1(g, v0, v1) is the unique solution of (2.6) and estimate (2.9) implies (2.7). Now let us consider
the last part of the proposition. For this purpose, let (g, v0, v1) ∈ H and let u ∈ H1(Q) be the solution of
(2.6). Note first that (∂2
t − ∆x)u = −qu ∈ L2(Q). Therefore, u ∈ H(Q) and τ1,1u ∈ H−3(0, T ; H−
3 2(∂Ω)), τ1,2u ∈ H−2(Ω) with kτ1,1uk2+ kτ1,2uk26 C2kuk2H (Q)= C 2(kuk2 H1(Q))+ kquk 2 L2(Q)) 6 C2(1 + kqk2Lp(Q)) kuk 2 H1(Q).
Combining this with (2.7), we find that Bq is a bounded operator from H to H−3(0, T ; H−
3
2(V0)) × H−2(Ω).
From now on, we define the set Cq(T, V ) by
Cq(T, V ) = {(g, v0, v1, Bq(g, v0, v1)) : (g, v0, v1) ∈ H}.
In the same way, for q ∈ Lp1(0, T ; Lp2(Ω)), p
1 > 2, p2 > n, we consider the set Cq(T ), Cq(0), Cq(0, T )
introduced before Theorem 1.1. Using similar arguments to Proposition 2.2 we can prove the following. Proposition 2.3. Let (g, v1) ∈ H∗ with v0= 0 and let q ∈ L∞(0, T ; Lp(Ω)), p > n. Then, the IBVP (2.6)
admits a unique weak solution u ∈ L2(0, T ; H1(Ω)) satisfying
kukL2(0,T ;H1(Ω))6 C k(g, v1)kH∗ (2.10)
and the boundary operator Bq,∗: (g, v1) 7→ (τ1,1u|V, τ1,2u) is a bounded operator from H∗ to
H−3(0, T ; H−32(V0)) × H−2(Ω).
We define the set Cq(0, T, V ) by
3. Proof of Theorem 1.1
The goal of this section is to prove Theorem 1.1. For this purpose, we consider special solutions uj of
the equation
∂t2uj− ∆xuj+ qjuj= 0 (3.1)
taking the form
uj = aj,1eiλψ1(t,x)+ aj,2eiλψ2(t,x)+ Rj,λ (3.2)
with a large parameter λ > 0 and a remainder term Rj,λ that admits some decay with respect to λ. The
use of such a solutions, also called oscillating geometric optics solutions, goes back to [40] who have proved unique recovery of time-independent coefficients. Since then, such approach has been used by various authors in different context including recovery of a bounded time-dependent coefficient by [34]. In this section we will prove how one can extend this approach, that has been specifically designed for the recovery of time-independent coefficients or bounded time-dependent coefficients, to the recovery of singular time-dependent coefficients.
3.1. Oscillating geometric optics solutions. Fixing ω ∈ Sn−1, λ > 1 and a
j,k ∈ C∞(Q), j = 1, 2,
k = 1, 2, we consider solutions of (3.1) taking the form
u1(t, x) = a1,1(t, x)e−iλ(t+x·ω)+ a1,2(t, x)e−iλ((2T −t)+x·ω)+ R1,λ(t, x), (t, x) ∈ Q, (3.3)
u2(t, x) = a2,1(t, x)eiλ(t+x·ω)+ a2,2(t, x)eiλ(−t+x·ω)+ R2,λ(t, x), (t, x) ∈ Q. (3.4)
Here, the expression aj,k, j, k = 1, 2, are independent of λ and they are respectively solutions of the transport
equation
∂taj,k+ (−1)kω · ∇xaj,k= 0, (t, x) ∈ Q, (3.5)
and the expression Rj,λ, j = 1, 2, solves respectively the IBVP
∂2 tR1,λ− ∆xR1,λ+ q1R1,λ = F1,λ, (t, x) ∈ Q, R1,λ(T, x) = 0, ∂tR1,λ(T, x) = 0, x ∈ Ω R1,λ(t, x) = 0, (t, x) ∈ Σ, (3.6) ∂t2R2,λ− ∆xR2,λ+ q2R2,λ = F2,λ, (t, x) ∈ Q, R2,λ(0, x) = 0, ∂tR2,λ(0, x) = 0, x ∈ Ω R2,λ(t, x) = 0, (t, x) ∈ Σ, (3.7)
with Fj,λ= −[(+qj)(uj−Rj,λ)]. The main point in the construction of such solutions, also called oscillating
geometric optics (GO in short) solutions, consists of proving the decay of the expression Rj,λwith respect
to λ → +∞. Actually, we would like to prove the following, lim
λ→+∞kRj,λkL∞(0,T ;L2(Ω))= 0. (3.8)
For q ∈ L∞(Q), the construction of GO solutions of the form (3.3)-(3.4), with aj,k satisfying (3.5) and Rj,σ
satisfying (3.6)-(3.8), has been proved in [34, Lemma 2.2]. The fact that q is bounded plays an important role in the arguments of [34, Lemma 2.2]. For this reason we can not apply the result of [34] and we need to consider the following.
Lemma 3.1. Let qj ∈ Lp1(0, T ; Lp2(Ω)), j = 1, 2, p1 > 2, p2 > n. Then, we can find uj ∈ K(Q) solving
(3.1), of the form (3.3)-(3.4), with Rj,λ, j = 1, 2, satisfying (3.8) and the following estimate
sup
λ>1
max
Proof. We will consider this result only for j = 2, the proof for j = 1 being similar by symmetry. Note first that, (3.5) implies that
F2,λ(t, x) = −eiλ(t+x·ω)( + q2)a2,1(t, x) − eiλ(−t+x·ω)( + q2)a2,2(t, x)
= Hλ(t, x)
= eiλtH1,λ(t, x) + e−iλtH2,λ(t, x),
with
kHλkL2(Q)6 k( + q2)a2,1kL2(Q)+ k( + q2)a2,2kL2(Q). (3.10)
Thus, in light Proposition 2.1, we have R2,λ∈ K(Q) with
kR2,λkC1([0,T ];L2(Ω))+ kR2,λkC([0,T ];H1(Ω))6 C(1 + kq2kLp1(0,T ;Lp2(Ω)))(ka2,1kW2,∞(Q)+ ka2,2kW2,∞(Q)).
(3.11) In particular, this proves (3.9). The only point that we need to check is the decay with respect to λ given by (3.8). For this purpose, we consider v(t, x) :=R0tR2,λ(s, x)ds and we easily check that v solves
∂2 tv − ∆xv = Gλ, (t, x) ∈ Q, v(0, x) = 0, ∂tv(0, x) = 0, x ∈ Ω v(t, x) = 0, (t, x) ∈ Σ, (3.12) with Gλ(t, x) = − Z t 0 q2(s, x)R2,λ(s, x)ds + Z t 0 Hλ(s, x)ds, (t, x) ∈ Q.
In view of [38, Theorem 2.1, Chapter 5], since Gλ∈ H1(0, T ; L2(Ω)) we have v ∈ H2(Q). We define the
energy E(t) at time t associated with v and given by
E(t) := Z Ω |∂tv|2(t, x) + |∇xv|2(t, x) dx > Z Ω |R2,λ(t, x)|2dx.
Multiplying (3.12) by ∂tv and taking the real part, we find
E(t) = −2R Z t 0 Z Ω Z s 0 q2(τ, x)R2,λ(τ, x)dτ ∂tv(s, x)dxds +2R Z t 0 Z Ω Z s 0 Hλ(τ, x)dτ ∂tv(s, x)dxds .
Applying Fubini’s theorem, we obtain
E(t) = −2R Z t 0 Z Ω q2(τ, x)R2,λ(τ, x) Z t τ ∂tv(s, x)ds ! dxdτ ! + 2R Z t 0 Z Ω Z s 0 Hλ(τ, x)dτ ∂tv(s, x)dxds = −2R Z t 0 Z Ω q2(τ, x)R2,λ(τ, x)(v(t, x) − v(τ, x))dxdτ + 2R Z t 0 Z Ω Z s 0 Hλ(τ, x)dτ ∂tv(s, x)dxds . (3.13) On the other hand, applying the Hölder inequality, we get
Z t 0 Z Ω q2(τ, x)R2,λ(τ, x)v(t, x)dxdτ 6 Z t 0 k∂tv(τ, ·)kL2(Ω)kq2(τ, ·)v(t, ·)kL2(Ω)dτ 6 Z t 0 k∂tv(τ, ·)kL2(Ω)kq2(τ, ·)kLp2(Ω)dτ kv(t, ·)k L p2−2 2p2 (Ω) .
Then, combining the Sobolev embedding theorem with the Poincarré inequality, we deduce that Z t 0 Z Ω q2(τ, x)R2,λ(τ, x)v(t, x)dxdτ 6 C Z t 0 k∂tv(τ, ·)kL2(Ω)kq2(τ, ·)kLp2(Ω)dτ kv(t, ·)kH1(Ω) 6 C Z t 0 E(τ )1/2kq2(τ, ·)kLp2(Ω)dτ E(t)1/2 6 C2 Z t 0 E(τ )1/2kq2(τ, ·)kLp2(Ω)dτ 2 +E(t) 4 6 C2T Z t 0 E(τ ) kq2(τ, ·)k2Lp2(Ω)dτ +E(t) 4 , with C depending only on Ω. Applying again the Hölder inequality, we get
Z t 0 Z Ω q2(τ, x)R2,λ(τ, x)v(t, x)dxdτ 6 C Z t 0 E(τ )p1−2p1 dτ p1−2p1 kq2k 2 Lp1(0,T ;Lp2(Ω))+ E(t) 4 . (3.14) In the same way, we obtain
Z t 0 Z Ω q2(τ, x)R2,λ(τ, x)v(τ, x)dxdτ 6 C Z t 0 E(τ ) kq2(τ, ·)kLp2(Ω)dτ 6 C Z t 0 E(τ )p1−2p1 dτ p1−2p1 kq2kLp1 2 (0,T ;Lp2(Ω)). (3.15) Finally, fixing βλ(t, x) := Z t 0 Hλ(τ, x)dτ, we find Z t 0 Z Ω Z s 0 Hλ(τ, x)dτ ∂tv(s, x)dxds 6 kβ λk Lp12 (0,T ;L2(Ω)) Z t 0 E(τ )2(p1−2)p1 dτ p1−2p1 6 kβλk 2 Lp12 (0,T ;L2(Ω))+ Z t 0 E(τ )2(p1−2)p1 dτ 2(p1−2)p1 6 kβλk 2 Lp12 (0,T ;L2(Ω))+ T p1−2 p1 Z t 0 E(τ )p1−2p1 dτ p1−2p1 . (3.16)
Combining (3.13)-(3.16), we deduce that
E(t) 6 E(t)4 + C(kq2kLp1(0,T ;Lp2(Ω))+ 1) 2 Z t 0 E(τ )p1−2p1 dτ p1−2p1 + kβλk 2 Lp12 (0,T ;L2(Ω)) and we get E(t) 6 C(kq2kLp1(0,T ;Lp2(Ω))+ 1) 2 Z t 0 E(τ )p1−2p1 dτ p1−2p1 + 4 kβλk2 Lp12 (0,T ;L2(Ω)) 3 ,
with C depending only on Ω and T . Now taking the power p1
p1−2 on both side of this inequality, we get
E(t)p1−2p1 6 2 p1 p1−2Cp1−2p1 (kq2kLp1(0,T ;Lp2(Ω))+ 1) 2p1 p1−2 Z t 0 E(τ )p1−2p1 dτ + 2p1−2p1 4 kβλk2 Lp12(0,T ;L2(Ω)) 3 p1 p1−2
and applying the Gronwall inequality, we obtain E(t)p1−2p1 6 C1kβλk 2p1 p1−2 Lp12 (0,T ;L2(Ω))e C2t 6 C1kβλk 2p1 p1−2 Lp12 (0,T ;L2(Ω))e C2T,
where C1depends only on p1and C2 on kq2kLp1(0,T ;Lp2(Ω)), Ω and T . According to this estimate, the proof
of the lemma will be completed if we prove that lim
λ→+∞kβλkL∞(0,T ;L2(Ω))= 0. (3.17)
This follows from some arguments similar to the end of the proof of [34, Lemma 2.2] that we recall for sake of completeness. Applying the Riemann-Lebesgue lemma, for all t ∈ [0, T ] and almost every x ∈ Ω, we have
lim λ→+∞ Z t 0 eiλτH1,λ(τ, x)dτ = lim λ→+∞ Z t 0 e−iλτH2,λ(τ, x)dτ = 0.
Therefore, for all t ∈ [0, T ] and almost every x ∈ Ω, we obtain lim
λ→+∞βλ(t, x) =λ→+∞lim
Z t
0
Hλ(τ, x)dτ = 0.
Moreover, from the definition of Hλ, we get
Z t 0 Hλ(τ, x)dτ 6 Z t 0 (|( + q2)a2,1| + |( + q2)a2,2|)ds, t ∈ [0, T ], x ∈ Ω.
Thus, we deduce from Lebesgue’s dominated convergence theorem that lim λ→+∞ Z t 0 Hλ(τ, ·)dτ L2(Ω) = 0, t ∈ [0, T ]. Combining this with the estimate
Z t2 0 Hλ(τ, ·)dτ − Z t1 0 Hλ(τ, ·)dτ L2(Ω) 6 (t2− t1) 1 2[k( + q2)a2,1k L2(Q)+ k( + q2)a2,2kL2(Q))], 0 6 t1< t26 T,
we deduce (3.17). This completes the proof the lemma. 3.2. Proof of Theorem 1.1 with restriction at t = 0 or t = T . In this section we will prove that (1.3) or (1.4) implies that q1 = q2. We start by assuming that (1.3) is fulfilled and we fix q = q2− q1 on
Q extended by 0 on R1+n\ Q. We fix λ > 1, ω ∈ Sn−1
and we fix ξ ∈ R1+n satisfying (1, −ω) · ξ = 0. Then, in view of Lemma 3.1, we can consider uj∈ K(Q), j = 1, 2, solving (3.1), of the form (3.3)-(3.4), with
a1,1(t, x) = (2π)
n+1
2 e−i(t,x)·ξ, a1,2= 0, a2,1= 1, a2,2 = −1 and with condition (3.8)-(3.9) fulfilled, that is,
u1(t, x) = (2π)
n+1
2 e−i(t,x)·ξe−iλ(t+x·w)+ R1,λ(t, x),
u2(t, x) = eiλ(t+x·w)− eiλ(−t+x·w)+ R2,λ(t, x).
(3.18)
Obviously, we have u2(0, x) = 0, since R2,λ(0, x) = 0 by (3.7). In view of Proposition 2.1, there exists a
unique weak solution v ∈ K(Q) to the IBVP:
∂t2v − ∆v + q1v = 0 in Q, v|t=0= u2|t=0= 0, ∂tv|t=0= ∂tu2|t=0, v|Σ= u2|Σ. Setting u := v − u2, we see ∂t2u − ∆u + q1u = (q2− q1)u2 in Q, u|t=0= 0, ∂tu|t=0= 0, u|Σ= 0. (3.19)
Noting that the inhomogeneous term (q2− q1)u2∈ L2(Q), due to the fact that q2− q1∈ L2(0, T ; Lp2(Ω)) and
u2∈ L∞(0, T ; H1(Ω)). Hence, again using Proposition 2.1 gives that u ∈ C([0, T ]; H01(Ω)) ∩ C1([0, T ]; L2(Ω)).
Therefore, we have u ∈ K(Q) ∩ H(Q). Combining this with u1∈ K(Q) ∩ H(Q), we deduce that
(∂tu, −∇xu), (∂tu1, −∇xu1) ∈ Hdiv(Q) = {F ∈ L2(Q; Cn+1) : div(t,x)F ∈ L2(Q)}.
Now, in view of [28, Lemma 2.2] we can multiply u1 to the equation in (3.19) and apply Green formula to
get Z Q (q2− q1)u2u1dxdt = Z Q (∂t2u − ∆u + q1u)u1dxdt = Z Q
(u + q1u)u1− (u1+ q1u1)u dxdt
= Z Q uu1− u1u dxdt = h(∂tu, −∇xu) · n, u1iH− 1 2(∂Q),H12(∂Q)− h(∂tu1, −∇xu1) · n, uiH− 12(∂Q),H12(∂Q) (3.20)
with n the outward unite normal vector to ∂Q. Since Cq1(0) = Cq2(0) and v|t=0 = u2|t=0 = 0, we see
∂νu|Σ= u|t=T = ∂tu|t=T = 0, in addition to the boundary conditions of u in (3.19). Consequently, it follows
from (3.20) that
Z
Q
(q2− q1)u2u1dxdt = 0.
Inserting the expressions of uj (j = 1, 2) given by (3.18) to the previous identity gives the relation
0 = (2π)(n+1)/2 Z Q q(t, x)e−i(t,x)·ξdxdt + ˜Rλ, ˜ Rλ:= (2π)(n+1)/2 Z Q
q(t, x)e−i(t,x)·ξ−e−2iλt+ e−iλ(t+x·w)R2,λ(t, x)
dxdt + Z Q q(t, x)R1,λ(t, x) eiλ(t+x·w)− eiλ(−t+x·w)+ R2,λ(t, x) dxdt
for all λ > 1. Using the fact that q ∈ L2(Q) and applying the Riemann-Lebesgue lemma and (3.8), we
deduce that Z Q
q(t, x)e−i(t,x)·ξ−e−2iλt+ e−iλ(t+x·w)R2,λ(t, x)
dxdt → 0, Z Q q(t, x)R1,λ(t, x) eiλ(t+x·w)− eiλ(−t+x·w)dxdt → 0 as λ → ∞. On the other hand, by Cauchy-Schwarz inequality it holds that
Z Q q(t, x)R1,λ(t, x)R2,λ(t, x) dxdt 6 ||q R 1,λ||L2(Q)||R2,λ||L2(Q) 6 C ||q||Lp1(0,T ;Lp2(Ω))||R1,λ||L∞(0,T ;H1(Ω))||R2,λ||L∞(0,T ;L2(Ω)),
which tends to zero as λ → ∞ due to the decaying behavior of Rj,λ(see (3.8)) and estimate (3.9). Therefore,
| ˜Rλ| → 0 as λ → ∞. It then follows that
F q(ξ) = (2π)n+12
Z
R1+n
q(t, x)e−i(t,x)·ξdxdt = 0. (3.21) Since ω ∈ Sn−1 is arbitrary chosen, we deduce that for any ω ∈ Sn−1 and any ξ lying in the hyperplane {ζ ∈ R1+n
q ∈ L1
(R1+n) is compactly supported in Q, we know that F q is a complex valued real-analytic function
and it follows that F q = 0. By inverse Fourier transform this yields the vanishing of q, which implies that q1= q2 in Q.
To prove that the relation (1.4) implies q1 = q2, we shall consider uj ∈ K(Q), j = 1, 2, solving (3.1),
of the form (3.3)-(3.4), with a1,1 = 1, a1,2 = −1, a2,1 = (2π)
n+1
2 e−i(t,x)·ξ, a2,2 = 0 and with condition
(3.8)-(3.9) fulfilled. Then, by using the fact that u1(T, x) = 0, x ∈ Ω, and by repeating the above arguments,
we deduce that q1= q2. For brevity we omit the details.
We have proved so far that either of the conditions (1.3) and (1.4) implies q1= q2. It remains to prove
that for T > Diam(Ω), the condition (1.5) implies q1= q2.
3.3. Proof of Theorem 1.1 with restriction at t = 0 and t = T . In this section, we assume that T > Diam(Ω) is fulfilled and we will show that (1.5) implies q1 = q2. For this purpose, we fix λ > 1,
ω ∈ Sn−1and ε = T −Diam(Ω)
4 . We set also χ ∈ C ∞
0 (−ε, T + Diam(Ω) + ε) satisfying χ = 1 on [0, T + Diam(Ω)]
and x0∈ Ω such that
x0· ω = inf x∈Ω
x · ω.
We introduce the solutions uj∈ K(Q), j = 1, 2, of (3.1), of the form (3.3)-(3.4), with
a1,1(t, x) = (2π) n+1 2 χ(t + (x − x0) · ω)e−i(t,x)·ξ, a1,2(t, x) = −(2π) n+1 2 χ((2T − t) + (x − x0) · ω)e−i(2T −t,x)·ξ, a2,1(t, x) = χ(t + (x − x0) · ω), a2,2(t, x) = −χ(−t + (x − x0) · ω)
and with condition (3.8)-(3.9) fulfilled. Then, one can check that u1(T, x) = u2(0, x) = 0, x ∈ Ω, and
repeating the arguments of the previous subsection we deduce that condition (1.5) implies the orthogonality identity
Z
Q
q(t, x)u2(t, x)u1(t, x)dxdt = 0. (3.22)
It remains to proves that this implies q = 0. Note that Z Q q(t, x)u2(t, x)u1(t, x)dxdt =(2π) n+1 2 Z R1+n q(t, x)χ2(t + (x − x0) · ω)e−i(t,x)·ξdxdt + Z Q e−2iλta1,1a2,2dxdt + Z Q e−2iλ(T −t)a1,2a2,1dxdt + e−2iλT Z Q a1,2a2,2dxdt + Z Q Zλ(t, x)dxdt, (3.23) with Zλ= q(u1− R1,λ)R2,λ+ q(u2− R2,λ)R1,λ+ qR2,λR1,λ.
In a similar way to the previous subsection, one can check that (3.8)-(3.9) imply that lim
λ→+∞
Z
Q
Zλdxdt = 0.
Moreover, the Riemann-Lebesgue lemma implies lim λ→+∞ Z Q e−2iλta1,1a2,2dxdt + Z Q e−2iλ(T −t)a1,2a2,1dxdt = 0. In addition, using the fact that for (t, x) ∈ Q we have
0 6 t + (x − x0) · ω 6 T + |x − x0| 6 T + Diam(Ω), we deduce that q(t, x)χ2(t + (x − x0) · ω) = q(t, x), (t, x) ∈ R1+n and that (2π)n+12 Z R1+n q(t, x)χ2(t + (x − x0) · ω)e−i(t,x)·ξdxdt = F q(ξ).
Thus, repeating the arguments of the previous subsection we can deduce that q1= q2provided that Z Q a1,2a2,2dxdt = 0. (3.24) Since a2,2(t, x) = −χ(−t + (x − x0) · ω) and a1,2 = −(2π) n+1
2 χ((2T − t) + (x − x0) · ω)e−i(t,x)·ξ, we deduce
that
supp(a2,2) ⊂ {(t, x) ∈ R1+n: (x − x0) · ω > t − ε},
supp(a1,2) ⊂ {(t, x) ∈ R1+n: 2T − t + (x − x0) · ω 6 T + Diam(Ω) + ε}.
But, for any (t, x) ∈ {(t, x) ∈ R1+n: (x − x0) · ω > t − ε}, one can check that
2T − t + (x − x0) · ω > 2T − ε = T + Diam(Ω) + 3ε > T + Diam(Ω) + ε.
Therefore, we have
{(t, x) ∈ R1+n: (x − x0) · ω > t − ε} ∩ {(t, x) ∈ R1+n: 2T − t + (x − x0) · ω 6 T + Diam(Ω) + ε} = ∅
and by the same way that supp(a2,2) ∩ supp(a1,2) = ∅. This implies (3.24) and by the same way that q1= q2.
Thus, the proof of Theorem 1.1 is completed.
4. Proof of Theorem 1.2
In the previous section we have seen that the oscillating geometric optics solutions (3.2) can be used for the recovery of some general singular time-dependent potential. We have even proved that, by adding a second term, we can restrict the data on the bottom t = 0 and top t = T of Q while avoiding a "reflection". Nevertheless, as mentioned in the introduction, it is not clear how one can adapt this approach to restrict data on the lateral boundary Σ without requiring additional smoothness or geometrical assumptions. In this section, we use a different strategy for restricting the data at Σ. Namely, we replace the oscillating GO solutions (3.2) by exponentially growing and decaying solutions, of the form (1.9), in order to restrict the data on Σ by mean of a Carleman estimate. In this section, we assume that q1, q2∈ Lp(Q), with p > n + 1,
and we will prove that (1.6) implies q1= q2. For this purpose, we will start with the construction of solutions
of (1.1) taking the form (1.9). Then we will show Carleman estimates for unbounded potentials and we will complete the proof of Theorem 1.2.
4.1. Geometric optics solutions for Theorem 1.2. Let ω ∈ Sn−1 and let ξ ∈ R1+n be such that ξ · (1, −ω) = 0. In this section we consider exponentially decaying solutions u1 ∈ H1(Q) of the equation
(∂2
t − ∆x+ q1)u1= 0 in Q taking the form
u1(t, x) = e−λ(t+x·ω)(e−i(t,x)·ξ+ w1(t, x)), (4.1)
and exponentially growing solution u2∈ H1(Q) of the equation (∂2t− ∆x+ q2)u2= 0 in Q taking the form
u2(t, x) = eλ(t+x·ω)(1 + w2(t, x)) (4.2)
where λ > 1 and the term wj∈ H1(Q), j = 1, 2, satisfies
kwjkH1(Q)+ λ kwjkL2(Q)6 C, (4.3)
with C independent of λ. We summarize these results in the following way.
Proposition 4.1. There exists λ1> 1 such that for λ > λ1 we can find a solution u1 ∈ H1(Q) of u1+
q1u1= 0 in Q taking the form (4.1) with w1∈ H1(Q) satisfying (4.3) for j = 1.
Proposition 4.2. There exists λ2 > λ1 such that for λ > λ2 we can find a solution u2 ∈ H1(Q) of
We start by considering Proposition 4.1. To build solutions u1∈ H1(Q) of the form (4.1), we first recall
some preliminary tools and a suitable Carleman estimate in Sobolev space of negative order borrowed from [33]. For all m ∈ R, we introduce the space Hm
λ (R
1+n) defined by
Hλm(R1+n) = {u ∈ S0(R1+n) : (|(τ, ξ)|2+ λ2)m2u ∈ Lˆ 2(R1+n)},
with the norm
kuk2Hm λ(R1+n)= Z R Z Rn (|(τ, ξ)|2+ λ2)m|ˆu(τ, ξ)|2dξdτ.
Note that here we consider these spaces with λ > 1 and, for λ = 1, one can check that Hλm(R1+n) = Hm(R1+n). Here for all tempered distribution u ∈ S0(R1+n), we denote by ˆu the Fourier transform of u. We fix the weighted operator
Pω,±λ:= e∓λ(t+x·ω)e±λ(t+x·ω)= ± 2λ(∂t− ω · ∇x)
and we recall the following Carleman estimate
Lemma 4.1. (Lemma 5.1, [33]) There exists λ01> 1 such that kvkL2(R1+n)6 C kPω,λvkH−1
λ (R1+n)
, v ∈ C0∞(Q), λ > λ01, (4.4) with C > 0 independent of v and λ.
From this result we can deduce the Carleman estimate
Lemma 4.2. Let p1∈ (n + 1, +∞), p2 ∈ (n, +∞) and q ∈ Lp1(Q) ∪ L∞(0, T ; Lp2(Ω)). Then, there exists
λ001 > λ01 such that
kvkL2(R1+n)6 C kPω,λv + qvkH−1 λ (R1+n)
, v ∈ C0∞(Q), λ > λ001, (4.5) with C > 0 independent of v and λ.
Proof. We start by considering the case q ∈ Lp1(Q). Note first that
kPω,λv + qvkH−1 λ (R1+n)> kPω,λ vkH−1 λ (R1+n)− kqvkH −1 λ (R1+n) . (4.6)
On the other hand, fixing
r = n + 1 p1 , 1 p3 = 1 2 − 1 p1 = n + 1 − 2r 2(n + 1) , by the Sobolev embedding theorem we deduce that
kqvkH−1 λ (R1+n)6 λ r−1 kqvkH−r λ (Rn+1) 6 λr−1kqvkH−r(Rn+1) 6 Cλr−1kqvk L p3 p3−1(Q)
Combining this with the fact that
p3− 1 p3 = 1 − 1 p3 = 1 2+ 1 p1 , we deduce from the Hölder inequality that
kqvkH−1
λ (R1+n)6 Cλ
r−1kqk
Lp1(Q)kvkL2(Q).
Thus, applying (4.4) and (4.6), we deduce (4.5) for λ > 1 sufficiently large. Now let us consider the case q ∈ L∞(0, T ; Lp2(Ω)). Note first that
kqvkH−1
λ (R1+n)6 kqvkL2(0,T ;H −1 λ (Rn))
. Therefore, by repeating the above arguments, we obtain
kqvkH−1
λ (R1+n)6 Cλ n p2−1kqk
which implies (4.5) for λ > 1 sufficiently large. Combining these two results, one can find λ001> λ01such that
(4.5) is fulfilled.
Using this new carleman estimate we are now in position to complete the proof of Proposition 4.1. Proof of Proposition 4.1. Note first that
(e−λ(t+x·ω)e−iξ·(t,x)) = [2iλ(1, −ω) · ξe−iξ·(t,x)+ e−iξ·(t,x)]e−λ(t+x·ω) = [e−iξ·(t,x)]e−λ(t+x·ω),
(e−λ(t+x·ω)w1) = e−λ(t+x·ω)Pω,−λw1.
Therefore, we need to consider w1∈ H1(Q) a solution of
Pω,−λw1+ q1w1= −eλ(t+x·ω)( + q1)(e−λ(t+x·ω)e−iξ·(t,x)) = −( + q1)e−iξ·(t,x)= F (4.7)
and satisfying (4.3) for j = 1. For this purpose, we will use estimate (4.5). From now on, we fix λ1 = λ001.
Applying the Carleman estimate (4.5), we define the linear form L on {Pω,λz + q1z : z ∈ C0∞(Q)}, considered
as a subspace of Hλ−1(R1+n) by L(Pω,λz + q1z) = Z Q zF dxdt, z ∈ C0∞(Q). Then, (4.5) implies |L(Pω,λz + q1z)| 6 C kF kL2(Q)kPω,λz + q1zkH−1 λ (R1+n) , z ∈ C0∞(Q).
Thus, by the Hahn Banach theorem we can extend L to a continuous linear form on Hλ−1(R1+n) still denoted
by L and satisfying kLk6 C kF kL2(Q). Therefore, there exists w1∈ Hλ1(R1+n) such that
hh, w1iH−1
λ (R1+n),H 1 λ(R1+n)
= L(h), h ∈ Hλ−1(R1+n).
Choosing h = Pω,λz + q1z with z ∈ C0∞(Q) proves that w1 satisfies Pω,−λw1+ q1w1= F in Q. Moreover, we
have kw1kH1
λ(R1+n)6 kLk 6 C kF kL2(Q). This proves that w1 fufills (4.3) which completes the proof of the
proposition.
Now let us consider the construction of the exponentially growing solutions given by Proposition 4.2. Combining [33, Lemma 5.4] with the arguments used in Lemma 4.2 we obtain the Carleman estimate. Lemma 4.3. There exists λ02> 0 such that for λ > λ02, we have
kvkL2(R1+n)6 C kPω,−λv + q2vkH−1 λ (R1+n)
, v ∈ C∞0 (Q), λ > λ03, (4.8) with C > 0 independent of v and λ.
In a similar way to Proposition 4.1, we can complete the proof of Proposition 4.2 by applying estimate (4.8).
4.2. Carleman estimates for unbounded potential. This subsection is devoted to the proof of a Car-leman estimate similar to [33, Theorem 3.1]. More precisely, we consider the following estimate.
Theorem 4.1. Let p1∈ (n+1, +∞), p2∈ (n, +∞) and assume that q ∈ Lp1(Q) (resp q ∈ L∞(0, T ; Lp2(Ω)))
and u ∈ C2(Q). If u satisfies the condition
then there exists λ3 > λ2 depending only on Ω, T and M > kqkLp1(Q) (resp M > kqkL∞(0,T ;Lp2(Ω))) such
that the estimate λR Ωe −2λ(T +ω·x)|∂ tu(T, x)|2dx + λRΣ +,ωe −2λ(t+ω·x)|∂ νu|2|ω · ν(x)| dσ(x)dt +R Qe −2λ(t+ω·x)[λ2|u|2 + |∂tu|2+ |∇xu|2]dxdt 6 CR Qe −2λ(t+ω·x) (∂2 t − ∆x+ q)u 2 dxdt + λ3R Ωe −2λ(T +ω·x)|u(T, x)|2 dx +CλR Ωe −2λ(T +ω·x)|∇ xu(T, x)|2dx + λRΣ−,ωe−2λ(t+ω·x)|∂νu|2|ω · ν(x)| dσ(x)dt (4.10)
holds true for λ> λ3 with C depending only on Ω, T and M .
Proof. Since the proof of this result is similar for q ∈ Lp1(Q) or q ∈ L∞(0, T ; Lp2(Ω)), we assume without
lost of generality that q ∈ Lp1(Q). Note first that for q = 0, (4.9) follows from [33, Theorem 3.1]. On the
other hand, we have e −λ(t+ω·x) (∂2t− ∆x+ q)u L2(Q)> e −λ(t+ω·x) (∂t2− ∆x)u L2(Q)− e −λ(t+ω·x)qu L2(Q)
and by the Hölder inequality we deduce that e −λ(t+ω·x) (∂t2− ∆x+ q)u L2(Q)> e −2λ(t+ω·x) (∂t2− ∆x)u L2(Q)− kqkLp1(Q) e −λ(t+ω·x)u Lp3(Q) with p3= p2p1 1−2. Now fix s := n+1
p1 ∈ (0, 1) and notice that
1 p3
=n + 1 − 2s 2(n + 1) . Thus, by the Sobolev embedding theorem, we have
e −λ(t+ω·x)u Lp3(Q)6 C e −λ(t+ω·x)u Hs(Q)
and by interpolation we deduce that e −λ(t+ω·x)u Lp3(Q)6 C Z Q e−2λ(t+ω·x)(λ2|u|2+ |∂ tu|2+ |∇xu|2dxdt s2 e −λ(t+ω·x)u 1−s L2(Q) 6 Cλs−1 Z Q e−2λ(t+ω·x)(λ2|u|2+ |∂ tu|2+ |∇xu|2dxdt 12 . On the other hand, in view of [33, Theorem 3.1], there exists λ0
3> 1 such that, for λ > λ03, we have
R Qe −2λ(t+ω·x)(λ2|u|2+ |∂ tu|2+ |∇xu|2)dxdt 6 CR Qe −2λ(t+ω·x) (∂2 t− ∆x)u 2 dxdt + λ3R Ωe −2λ(T +ω·x)|u(T, x)|2 dx +CλR Ωe −2λ(T +ω·x)|∇ xu(T, x)| 2 dx + λR Σ−,ωe −2λ(t+ω·x)|∂ νu| 2 |ω · ν(x)| dσ(x)dt. Thus, we get R Qe −2λ(t+ω·x) (∂t2− ∆x+ q)u 2 dxdt + λ3RΩe−2λ(T +ω·x)|u(T, x)|2dx +λR Ωe −2λ(T +ω·x)|∇ xu(T, x)| 2 dx + λR Σ−,ωe −2λ(t+ω·x)|∂ νu| 2 |ω · ν(x)| dσ(x)dt >12 R Qe −2λ(t+ω·x) (∂2 t− ∆x)u 2 dxdt + λ3R Ωe −2λ(T +ω·x)|u(T, x)|2 dx +λR Ωe −2λ(T +ω·x)|∇ xu(T, x)| 2 dx + λR Σ−,ωe −2λ(t+ω·x)|∂ νu| 2 |ω · ν(x)| dσ(x)dt −C kqk2Lp1(Q)λ2(s−1)R Qe −2λ(t+ω·x) (∂2 t − ∆x)u 2 dxdt + λ3R Ωe −2λ(T +ω·x)|u(T, x)|2 dx −C kqk2Lp1(Q)λ 2(s−1)λR Ωe −2λ(T +ω·x)|∇ xu(T, x)| 2 dx + λR Σ−,ωe −2λ(t+ω·x)|∂ νu| 2 |ω · ν(x)| dσ(x)dt. Therefore, fixing λ sufficiently large and applying [33, Theorem 3.1] with a = q = 0 we deduce (4.10).
Remark 4.1. Note that, by density, (4.10) remains true for u ∈ C1([0, T ]; L2(Ω))∩C([0, T ]; H1(Ω)) satisfying
(4.9), (∂2
t − ∆x)u ∈ L2(Q) and ∂νu ∈ L2(Σ).
4.3. Completion of the proof of Theorem 1.2. This subsection is devoted to the proof of Theorem 1.2. From now on, we set q = q2− q1 on Q and we assume that q = 0 on R1+n\ Q. For all θ ∈ Sn−1 and all
r > 0, we set
∂Ω+,r,θ = {x ∈ ∂Ω : ν(x) · θ > r}, ∂Ω−,r,θ = {x ∈ ∂Ω : ν(x) · θ 6 r}
and Σ±,r,θ = (0, T ) × ∂Ω±,r,θ. Here and in the remaining of this text we always assume, without mentioning
it, that θ and r are chosen in such way that ∂Ω±,r,±θ contain a non-empty relatively open subset of ∂Ω.
Without lost of generality we assume that there exists ε > 0 such that for all ω ∈ {θ ∈ Sn−1: |θ − ω 0| 6 ε}
we have ∂Ω−,ε,ω⊂ V0. In order to prove Theorem 1.2, we will use the Carleman estimate stated in Theorem
4.1. Let λ > λ3 and fix ω ∈ {θ ∈ Sn−1: |θ − ω0| 6 ε}. According to Proposition 4.1, we can introduce
u1(t, x) = e−λ(t+x·ω)(e−i(t,x)·ξ+ w1(t, x)), (t, x) ∈ Q,
where u1∈ H1(Q) satisfies ∂t2u1− ∆xu1+ q1u1= 0, ξ · (1, −ω) = 0 and w1satisfies (4.3) for j = 1. Moreover,
in view of Proposition 4.2, we consider u2∈ H1(Q) a solution of ∂t2u2− ∆xu2+ q2u2= 0, of the form
u2(t, x) = eλ(t+x·ω)(1 + w2(t, x)), (t, x) ∈ Q,
where w2satisfies (4.3) for j = 2. In view of Proposition 2.2, there exists a unique weak solution z1∈ H(Q)
of ∂2 tz1− ∆xz1+ q1z1= 0 in Q, τ0z1= τ0u2. (4.11) Then, u = z1− u2 solves ∂2 tu − ∆xu + q1u = (q2− q1)u2 in Q, u(0, x) = ∂tu(0, x) = 0 on Ω, u = 0 on Σ. (4.12)
Since u2 ∈ H1(Q), by the Sobolev embedding theorem we have (q2− q1)u2 ∈ L2(Q). Thus, repeating the
arguments of Theorem 1.1, we derive the formula (3.20). On the other hand, we have u|t=0= ∂tu|t=0= u|Σ=
0 and condition (1.6) implies that u|t=T = ∂νu|V = 0. In addition, in view of [36, Theorem 2.1], we have
∂νu ∈ L2(Σ). Combining this with the fact that u ∈ C1([0, T ]; L2(Ω)) and u1 ∈ H1(Q) ⊂ H1(0, T ; L2(Ω)),
we obtain Z Q qu2u1dxdt = − Z Σ\V ∂νuu1dσ(x)dt + Z Ω ∂tu(T, x)u1(T, x)dx. (4.13)
Applying the Cauchy-Schwarz inequality to the first expression on the right hand side of this formula and using the fact that (Σ \ V ) ⊂ Σ+,ε,ω, we get
Z Σ\V ∂νuu1dσ(x)dt 6 Z Σ+,ε,ω ∂νue −λ(t+ω·x)(e−i(t,x)·ξ+ w 1) dσ(x)dt 6 C(1 + kw1kL2(Σ)) Z Σ+,ε,ω e −λ(t+ω·x)∂ νu 2 dσ(x)dt !12 , for some C independent of λ. On the other hand, one can check that
kw1kL2(Σ))6 C kw1kH1(Q).
Combining this with (4.3), we obtain Z Σ\V ∂νuu1dσ(x)dt 6 C Z Σ+,ε,ω e −λ(t+ω·x)∂ νu 2 dσ(x)dt !12 .
In the same way, we have Z Ω ∂tu(T, x)u1(T, x)dx 6 Z Ω ∂tu(T, x)e −λ(T +ω·x)e−i(T,x)·ξ+ w 1(T, x) dx 6 C Z Ω e −λ(T +ω·x)∂ tu(T, x) 2 dx 12 .
Combining these estimates with the Carleman estimate (4.10) and applying the fact that u|t=T = ∂νu|Σ−,ω =
0, ∂Ω+,ε,ω ⊂ ∂Ω+,ω, we find Z Q qu2u1dxdt 2 6 C Z Σ+,ε,ω e −λ(t+ω·x)∂ νu 2 dσ(x)dt + Z Ω e −λ(T +ω·x)∂ tu(T, x) 2 dx ! 6 ε−1Cλ−1 λ Z Σ+,ω e −λ(t+ω·x)∂ νu 2 ω · ν(x)dσ(x)dt + λ Z Ω e −λ(T +ω·x)∂ tu(T, x) 2 dx ! 6 ε−1Cλ−1 Z Q e −λ(t+ω·x)(∂2 t− ∆x+ q1)u 2 dxdt 6 ε−1Cλ−1 Z Q e −λ(t+ω·x)qu 2 2 dxdt 6 ε−1Cλ−1 Z Q |q(1 + w2)| 2 dxdt . (4.14)
Here C > 0 stands for some generic constant independent of λ. On the other hand, in a similar way to Lemma 4.2, combining the Hölder inequality and the Sobolev embedding theorem we get
Z Q |q(1 + w2)| 2 dxdt 6 C kqk2Lp(Q)k(1 + w2)k 2 H n+1 p (Q)6 C kqk 2 Lp(Q)(1 + kw2kH1(Q)) 2.
Combining this with (4.3) and (4.14), we obtain Z Q qu2u1dxdt 6 Cλ −1/2. It follows lim λ→+∞ Z Q qu2u1dxdt = 0. (4.15) Moreover, (4.1)-(4.3) imply Z Q qu2u1dxdt = Z R1+n q(t, x)e−iξ·(t,x)dxdt + Z Q Wλ(t, x)dxdt, with Z Q |Wλ(t, x)|dxdt 6 Cλ−1.
Combining this with (4.15), for all ω ∈ {y ∈ Sn−1 : |y − ω
0| 6 ε} and all ξ ∈ (1, −ω)⊥ := {ζ ∈ R1+n :
ζ ·(1, −ω) = 0}, the Fourier transform F (q) of q satisfies F (q)(ξ) = 0. On the other hand, since q ∈ L1
(R1+n)
is supported onQ which is compact, F (q) is a complex valued real-analytic function and it follows that q = 0 and q1= q2. This completes the proof of Theorem 1.2.
5. Proof of Theorem 1.3
Let us first remark that, in contrast to Theorem 1.1, in Theorem 1.2 we do not restrict the data to solutions of (1.1) satisfying u|t=0= 0. In this section we will show Theorem 1.3 by combining the restriction on the bottom t = 0, the top t = T of Q stated in Theorem 1.1 with the restriction on the lateral boundary Σ stated in Theorem 1.2. From now on, we fix q1, q2 ∈ L∞(0, T ; Lp(Ω)), p > n, and we will show that
condition (1.7) implies q1 = q2. For this purpose we still consider exponentially growing and decaying GO
solutions close to those of the previous subsection, but this time we need to take into account the constraint u2(0, x) = 0 required in Theorem 1.3. For this purpose, we will consider a different construction comparing
to the one of the previous section which will follow from a Carleman estimate in negative order Sobolev space only with respect to the space variable.
5.1. Carleman estimate in negative Sobolev space for Theorem 1.3. In this subsection we will derive a Carleman estimate in negative order Sobolev space which will be one of the main tool for the construction of exponentially growing solutions u2 of (3.1) taking the form
u2(t, x) = eλ(t+x·ω)(1 + w2(t, x)) − eλ(−t+x·ω), (5.1)
with the restriction τ0,2u2 = 0 (recall that for v ∈ C∞(Q), τ0,2v = v|t=0). In a similar way to the previous
section, for all m ∈ R, we introduce the space Hλm(R
n) defined by
Hλm(R n
) = {u ∈ S0(Rn) : (|ξ|2+ λ2)m2u ∈ Lˆ 2(Rn)},
with the norm
kuk2Hm λ(Rn)=
Z
Rn
(|ξ|2+ λ2)m|ˆu(ξ)|2dξ.
In order to construct solutions u2 of the form (4.2) and satisfying τ0,2u2 = 0, instead of the Carleman
estimate (4.8), we consider the following.
Theorem 5.1. There exists λ02> 0 such that for λ > λ02 and for all v ∈ C2([0, T ]; C∞
0 (Ω)) satisfying v(T, x) = ∂tv(T, x) = v(0, x) = 0, x ∈ Rn, (5.2) we have kvkL2((0,T )×Rn)6 C kPω,−λv + q2vkL2(0,T ;H−1 λ (Rn)) , (5.3)
with C > 0 independent of v and λ.
In order to prove this theorem, we start by recalling the following intermediate tools. From now on, for m ∈ R and ξ ∈ Rn, we set
hξ, λi = (|ξ|2+ λ2)12
and hDx, λimu defined by
hDx, λi m
u = F−1(hξ, λimF u). For m ∈ R we define also the class of symbols
Sλm= {cλ∈ C∞(Rn× Rn) : |∂αx∂ β
ξcλ(x, ξ)| 6 Cα,βhξ, λi m−|β|
, α, β ∈ Nn}.
Following [24, Theorem 18.1.6], for any m ∈ R and cλ∈ Sλm, we define cλ(x, Dx), with Dx= −i∇x, by
cλ(x, Dx)u(x) = (2π)− n 2 Z Rn cλ(x, ξ)ˆu(ξ)eix·ξdξ.
For all m ∈ R, we define OpSmλ := {cλ(x, Dx) : cλ∈ Sλm} and for m = −∞ we set
OpSλ−∞:= \
m∈R
OpSλm.
Lemma 5.1. There exists λ002 > 0 such that for λ > λ002 and for all v ∈ C2([0, T ]; C0∞(Ω)) satisfying (5.2), we
have
kvkL2(0,T ;H1
λ(Rn)) 6 C kPω,−λvkL2((0,T )×Rn)), (5.4)
with C > 0 independent of v and λ.
Proof. Consider w(t, x) = v(T − t, x) and note that according to (5.2), we have w ∈ C2([0, T ]; C∞
0 (Ω)) and
w(0, x) = ∂tw(0, x) = w(T, x) = 0, x ∈ Rn.
Therefore, in a similar way to the proof of [32, Lemma 4.1], one can check that Z Q |P−ω,λw|2dxdt > Z Q |w|2dxdt + cλ2Z Q |w|2dxdt, (5.5)
with c > 0 independent of w and λ. Now, recalling that w solves ∂t2w − ∆xw = w, (t, x) ∈ Q, w(0, x) = 0, ∂tw(0, x) = 0, x ∈ Ω w(t, x) = 0 (t, x) ∈ Σ, we deduce that Z Q |∇xw|2dxdt 6 C Z Q |w|2dxdt,
where C depends only on T and Ω. Combining this with (5.5), we obtain kwk2L2(0,T ;H1
λ(Rn))6 C
Z
Q
|P−ω,λw|2dxdt.
Using the fact that P−ω,λw(t, x) = Pω,−λv(T − t, x), we deduce (5.4).
Armed with this Carleman estimate, we are now in position of completing the proof of Theorem 5.1. Proof of Theorem 5.1. Let v ∈ C2([0, T ]; C0∞(Ω)) satisfy (5.2), consider Ωj, j = 1, 2, two bounded open
smooth domains of Rn such that Ω ⊂ Ω1, Ω1 ⊂ Ω2 and let ψ ∈ C0∞(Ω2) be such that ψ = 1 on Ω1. We
consider w ∈ C2([0, T ]; C∞
0 (Ω)) given by
w(t, ·) = ψ hDx, λi −1
v(t, ·) and we remark that w satisfies
w(T, x) = ∂tw(T, x) = w(0, x) = 0, x ∈ Rn. (5.6)
Now let us consider the quantity hDx, λi−1Pω,−λhDx, λi w. Note first that
kPω,−λhDx, λi wkL2(0,T ;H−1 λ (Rn)) = hDx, λi −1 Pω,−λhDx, λi w L2((0,T )×Rn).
Moreover, it is clear that
hDx, λi −1 Pω,−λhDx, λi = Pω,−λ. Therefore, we have kPω,−λhDx, λi wkL2(0,T ;H−1 λ (Rn)) = kPω,−λwkL2((0,T )×Rn)
and, since w satisfies (5.6), combining this with (5.4) we deduce that kwkL2(0,T ;H1
λ(Rn))6 C kPω,−λhDx, λi wkL2(0,T ;H −1 λ (Rn))
. (5.7)
On the other hand, fixing ψ1∈ C0∞(Ω1) satisfying ψ1= 1 on Ω, we get
w(t, ·) = hDx, λi −1
v(t, ·) + (ψ − 1) hDx, λi −1
and, combining this with (5.7), we deduce that kvkL2((0,T )×Rn) = hDx, λi −1 v L2(0,T ;H1 λ(Rn)) 6 kwkL2(0,T ;H1 λ(Rn))+ (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H1 λ(Rn)) 6 C kPω,−λhDx, λi wkL2(0,T ;H−1 λ (Rn)) + (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H1 λ(Rn)) 6 C kPω,−λvkL2(0,T ;H−1 λ (Rn)) + C Pω,−λhDx, λi (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H−1 λ (Rn)) + (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H1 λ(Rn)) . (5.8)
Moreover, since (ψ − 1) = 0 on neighborhood of supp(ψ1), in view of [24, Theorem 18.1.8], we have (ψ −
1) hDx, λi −1
ψ1∈ OpSλ−∞. In the same way, [24, Theorem 18.1.8] implies that
Pω,−λhDx, λi (ψ − 1) hDx, λi −1
ψ1∈ OpSλ−∞
and we deduce that C Pω,−λhDx, λi (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H−1 λ (Rn)) + (ψ − 1) hDx, λi −1 ψ1v L2(0,T ;H−1 λ (Rn)) 6 kvkL2 ((0,T )×Rn) λ2 .
Combining this with (5.8) and choosing λ sufficiently large, we deduce (5.3) for q2 = 0. Then, we deduce
(5.3) for q26= 0 by applying arguments similar to Lemma 4.2.
Applying the Carleman estimate (5.3), we can now build solutions u2 of the form (5.1) and satisfying
τ0,2u2= 0 and complete the proof of Theorem 1.3.
5.2. Completion of the proof of Theorem 1.3. We start by proving existence of a solution u2 ∈
L2(0, T ; H1(Ω)) of the form (5.1) with the term w
2∈ L2(0, T ; H1(Ω)) ∩ e−λ(t+x·ω)H,∗(Q), satisfying
kw2kL2(0,T ;H1(Ω))+ λ kw2kL2(Q)6 C, (5.9)
τ0,2w2= 0. (5.10)
This result is summarized in the following way.
Proposition 5.1. There exists λ2> λ1 such that for λ > λ2 we can find a solution u2 ∈ L2(0, T ; H1(Ω))
of u2+ q2u2 = 0 in Q taking the form (5.1) with w2 ∈ L2(0, T ; H1(Ω)) ∩ e−λ(t+x·ω)H,∗(Q) satisfying
(5.9)-(5.10).
Proof. We need to consider w2∈ L2(0, T ; H1(Ω)) a solution of
Pω,λw2+ q2w2= −e−λ(t+x·ω)( + q2)(eλ(t+x·ω)− eλ(−t+x·ω)) = −q2(1 − e−2λt), (5.11)
satisfying (5.9)-(5.10). Note that here, we use (5.11) and the fact that Pω,λw2= e−λ(t+x·ω)eλ(t+x·ω)w2 in
order to prove that w2 ∈ e−λ(t+x·ω)H,∗(Q) and we define τ0,2w2 by τ0,2w2 = e−λx·ωτ0,2eλ(t+x·ω)w2. We
will construct such a function w2 by applying estimate (5.3). From now on, we fix λ2 = λ002. Applying the
Carleman estimate (5.3), we define the linear form M on
I = {Pω,−λv + q2v : v ∈ C2([0, T ]; C0∞(Ω)) satisfying (5.2)}, considered as a subspace of L2(0, T ; Hλ−1(Rn)), by M(Pω,−λv + q2v) = − Z Q vq2(1 − e−2λt)dxdt, v ∈ I. Then, (5.3) implies |M(Pω,−λv + q1v)| 6 C kq2kL2(Q)kPω,−λv + q1vkL2(0,T ;H−1 λ (Rn) , v ∈ I,
with C > 0 independent of λ and v. Thus, by the Hahn Banach theorem we can extend M to a continuous linear form on L2(0, T ; H−1
λ (R n
)) still denoted by M and satisfying kMk 6 C kq2kL2(Q). Therefore, there
exists w2∈ L2(0, T ; Hλ1(Rn)) such that
hg, w2iL2(0,T ;H−1
λ (Rn),L2(0,T ;H 1 λ(Rn))
= M(g), g ∈ L2(0, T ; Hλ−1(Rn).
Choosing g = Pω,−λv + q2v with v ∈ C0∞(Q) proves that w2 satisfies Pω,λw2+ q2w1= −q2(1 − e−2λt) in Q.
Moreover, choosing g = Pω,−λv + q1v, with v ∈ I and ∂tv|t=0arbitrary, proves that (5.10) is fulfilled. Finally,
using the fact that kw2kL2(0,T ;H1
λ(Rn))6 kMk 6 C kqkL2(Q)proves that w2fulfills (5.9) which completes the
proof of the proposition.
Using this proposition, we are now in position to complete the proof of Theorem 1.3.
Proof of Theorem 1.3. Let us remark that since Lemma 4.2 and Theorem 4.1 are valid when q ∈ L∞(0, T ; Lp(Ω)) one can easily extend Proposition 4.1 to the case q
1∈ L∞(0, T ; Lp(Ω)). Therefore, in the
context of this section, Proposition 4.1 holds true. Combining Proposition 4.1 with Proposition 5.1, we deduce existence of a solution u1∈ H1(Q) of u1+ q1u1= 0 in Q taking the form (4.1), with w1∈ H1(Q)
satisfying (4.3) for j = 1, as well as the existence of a solution u2 ∈ L2(0, T ; H1(Ω)) of u2+ q2u2 = 0
in Q, τ0,2u2 = 0, taking the form (5.1) with the term w2 ∈ L2(0, T ; H1(Ω)) satisfying (5.9). Repeating
the arguments of the end of the proof of Theorem 1.2 (see Subsection 4.4), we can deduce the following orthogonality identity lim λ→+∞ Z Q qu1u2dxdt = 0. (5.12)
Moreover, one can check that Z Q qu1u2dxdt = Z R1+n q(t, x)e−iξ·(t,x)dxdt + Z Q Yλ(t, x)dxdt, with
Yλ(t, x) = q[e−2λte−i(t,x)·ξ+ e−i(t,x)·ξw2+ w1+ w1w2].
Combining (4.3), (5.9) with the fact that Z Q |q(t, x)| e −2λte−i(t,x)·ξ dxdt 6 kqkL2(Q) Z +∞ 0 e−2λtdt 12 6 kqkL2(Q)λ −1 2, we deduce that lim λ→+∞ Z Q Yλ(t, x)dxdt = 0.
Combining this asymptotic property with (5.12), we can conclude in a similar way to Theorem 1.2 that
q1= q2.
Acknowledgments
The work of the first author is supported by the NSFC grant (No. 11671028), NSAF grant (No. U1530401) and the 1000-Talent Program of Young Scientists in China. The second author would like to thank Pedro Caro for his remarks and fruitful discussions about this problem. The second author would like to thank the Beijing Computational Science Research Center, where part of this article was written, for its kind hospitality
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